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### Problem 1: Tension Member Design and Strength Calculations
The question asks for the design and allowable strengths for a tension member in both LRFD (Load and Resistance Factor Design) and ASD (Allowable Strength Design), considering both tensile yielding and tensile rupture. The given data are:
- Plate dimensions: 3/8" x 7" 
- Bolt diameter: 1 inch
- Steel grade: Fy = 36 ksi, Fu = 58 ksi
- Net area \( A_e = A \) (assumed that effective net area is equal to gross area)

#### (a) Design Strength for LRFD – Tensile Yielding
For LRFD, the design strength based on yielding is given by:
\[
\phi T_y = 0.9 \cdot A_g \cdot F_y
\]
Where:
- \( \phi = 0.9 \) is the resistance factor for yielding,
- \( A_g \) is the gross area of the plate, calculated as \( A_g = \text{thickness} \times \text{width} = 3/8" \times 7" \),
- \( F_y = 36 \text{ ksi} \).

#### (b) Allowable Strength for ASD – Tensile Yielding
For ASD, the allowable tensile strength based on yielding is:
\[
T_y = \frac{A_g \cdot F_y}{\Omega}
\]
Where:
- \( \Omega = 1.67 \) is the safety factor for yielding.

#### (c) Design Strength for LRFD – Tensile Rupture
For rupture, the LRFD strength is calculated as:
\[
\phi T_u = 0.75 \cdot A_e \cdot F_u
\]
Where:
- \( \phi = 0.75 \) is the resistance factor for rupture,
- \( A_e \) is the net area of the plate (assuming gross area in this case),
- \( F_u = 58 \text{ ksi} \).

#### (d) Allowable Strength for ASD – Tensile Rupture
For ASD, the allowable tensile strength based on rupture is:
\[
T_u = \frac{A_e \cdot F_u}{\Omega}
\]
Where:
- \( \Omega = 2.0 \) is the safety factor for rupture.

### Problem 2: Nominal Block Shear Strength
For block shear, the nominal strength is calculated using two potential failure modes: shear yielding on the gross area and tensile rupture on the net area.

#### Given Data:
- \( F_y = 276 \text{ MPa} \)
- \( F_u = 400 \text{ MPa} \)
- Bolt diameter: \( 7/8" \)

The block shear strength \( P_{bs} \) is calculated as:
\[
P_{bs} = 0.6 \cdot F_u \cdot A_{nv} + F_y \cdot A_{gt}
\]
Where:
- \( A_{nv} \) is the net area in shear,
- \( A_{gt} \) is the gross area in tension.

#### Steps:
1. Calculate the net area in shear: \( A_{nv} = ( \text{width} - \text{bolt diameter}) \times \text{thickness} \).
2. Calculate the gross area in tension: \( A_{gt} = \text{length of failure path in tension} \times \text{thickness} \).
3. Use these values to find the nominal block shear strength.

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Would you like the full calculation details for any of the specific parts? 

Here are five related questions:
1. How do you calculate the gross and net areas for tensile rupture?
2. What is the difference between LRFD and ASD design approaches?
3. How does the bolt hole size affect the net area in tension?
4. What safety factors are used in ASD for different failure modes?
5. Can you explain block shear failure in more detail?

**Tip**: Always double-check your units when dealing with mixed systems (e.g., ksi and MPa) to avoid calculation errors.

A PL 3/8 x 7 tension member is connected with three 1-inch-diameter bolts. The steel is Fy = 36 ksi, Fu = 58 ksi. Assume that Ae = Ag, and compute the design strength for LRFD consider tensile yielding, allowable strength for ASD consider tensile yielding, the design strength for LRFD consider tensile rupture, and the design strength for ASD consider tensile rupture. Also compute the nominal block shear strength of the tension member shown in Figure P3.5-1. Fy = 276 MPa, Fu = 400 MPa, with bolts 7/8 inch in diameter.

Learn how to compute the design and block shear strength for a PL 3/8 x 7 tension member connected by 1-inch diameter bolts, using steel with Fy = 36 ksi and Fu = 58 ksi. The process includes calculations for LRFD and ASD, tensile yielding, and tensile rupture, along with the nominal block shear strength for a member with Fy = 276 MPa and Fu = 400 MPa.

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